22v^2=28

Solution for 22v^2=28 equation:

22v^2=28

We move all terms to the left:

22v^2-(28)=0

a = 22; b = 0; c = -28;
Δ = b2-4ac
Δ = 02-4·22·(-28)
Δ = 2464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2464}=\sqrt{16*154}=\sqrt{16}*\sqrt{154}=4\sqrt{154}$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{154}}{2*22}=\frac{0-4\sqrt{154}}{44}
=-\frac{4\sqrt{154}}{44}
=-\frac{\sqrt{154}}{11}
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{154}}{2*22}=\frac{0+4\sqrt{154}}{44}
=\frac{4\sqrt{154}}{44}
=\frac{\sqrt{154}}{11}
$